\(a,f'\left(x\right)=3x^2-6x\\ f'\left(x\right)\le0\Leftrightarrow3x^2-6x\le0\\ \Leftrightarrow3x\left(x-2\right)\le0\Leftrightarrow0\le x\le2\)

Lời giải:

a. $f'(x)\leq 0$

$\Leftrightarrow 3x^2-6x\leq 0$

$\Leftrightarrow x(x-2)\leq 0$

$\Leftrightarrow 0\leq x\leq 2$

b.

$f'(x)=x^2-3x+2=0$

$\Leftrightarrow 3x^2-6x=x^2-3x+2=0$

$\Leftrightarrow 3x(x-2)=(x-1)(x-2)=0$

$\Leftrightarrow x-2=0$

$\Leftrightarrow x=2$

c.

$g(x)=f(1-2x)+x^2-x+2022$

$g'(x)=(1-2x)'f(1-2x)'_{1-2x}+2x-1$

$=-2[3(1-2x)^2-6(1-2x)]+2x-1$
$=-24x^2+2x+5$

$g'(x)\geq 0$

$\Leftrightarrow -24x^2+2x+5\geq 0$

$\Leftrightarrow (5-12x)(2x-1)\geq 0$

$\Leftrightarrow \frac{-5}{12}\leq x\leq \frac{1}{2}$

4: =>2x-3>5 hoặc 2x-3<-5

=>x>4 hoặc x<-1

5: =>-4<=2x-1<=4

=>-3/2<=x<=5/2

Vì $3x^2-x+1>0,x^2+1>0$

$\to \begin{cases}x^2 \geq 4\x<-1\\\end{cases}$

$\to \begin{cases}\left[ \begin{array}{l}x \geq 2\\x \leq -2\end{array} \right.\\x<-1\\\end{cases}$

$\to x \leq -2$

Vậy tập xác định của phương trình là `(-oo,-2]`

a. TH1:

\(\left\{{}\begin{matrix}x^2+3x-4< 0\\3-2x>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 1\\x>-4\end{matrix}\right.\\x>\dfrac{3}{2}\end{matrix}\right.\)

TH2:

\(\left\{{}\begin{matrix}x^2+3x-4>0\\3-2x< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1\\x< -4\end{matrix}\right.\\x< \dfrac{3}{2}\end{matrix}\right.\)

Vậy nghiệm của BPT:

\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 1\\x>-4\end{matrix}\right.\\x>\dfrac{3}{2}\end{matrix}\right.\)      \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1\\x< -4\end{matrix}\right.\\x< \dfrac{3}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left(x^2+2x\right)\left(3x^2+4x+8\right)>0\)

\(\Leftrightarrow\left(x>-2/x\ne0\right)\)

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